\(n_{H_2S}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(H_2S+\dfrac{3}{2}O_2\underrightarrow{^{^{t^0}}}SO_2+H_2O\)
\(0.4.................0.4\)
\(n_{NaOH}=n_{SO_2}=0.4\left(mol\right)\)
=> Tạo muối axit
\(NaOH+SO_2\rightarrow NaHSO_3\)
\(0.4............................0.4\)
\(m_{NaHSO_3}=0.4\cdot104=41.6\left(g\right)\)